Assume we have a large supply of 1, 2 and 3 ohm electrical resistors. If we sele
ID: 3277860 • Letter: A
Question
Assume we have a large supply of 1, 2 and 3 ohm electrical resistors. If we select three of them at random (each value equally likely). 3. (a) What is the probability the three resistor values add up to exactly 6 ohms? Show the actual combinations. Answer b) What is the probability the three resistor values add up to at least 6 ohms? Show the actual combinations. Answer 4. A system includes three components (A, B and C), one of which will fail over a time period. The probabilities of the mutually exclusive component failures are P(A) = 0.5 P(B)-0.4 P(C)-0.1 The probability of a system failure (F) given each component failure is P(HA) = 0.05 P(FB)-0.1 PFC-0.2 (a) Find the total probability, P(F), that the system will fail in the time period. P(F)- (b)If the system has failed, what are the probabilities that the failure was due to A, B and C respectively? Use Bayes' formula. P(AF) POF)Explanation / Answer
3. a) The probability the three resistor values add upto exactly 6 ohms is 2/27
since
Sample Space = 3*3*3 = 27 = n
Let E denote sum is 6
i.e. E={(1,2,3), (2,2,2)}, m = 2
P(E) = 2/27
b) The probability the three resistor values add upto exactly 6 ohms is 9/27
E = { (1,2,3), (2,2,2), (2,2,3), (2,3,2), (3,2,2), (3,3,2), (3,2,3), (2,3,3), (3,3,3) } m = 9
P(E) = 9/27
4.
a) P(F) = P(A)P(F/A) + P(B)P(F/B) + P(C)P(F/C)
= 0.5*0.05 + 0.4*0.1 + 0.1*0.2 = 0.085
b) P(A/F) = P(A) P(F/A) / P(F)
= 0.5*0.05 / 0.085= 0.2941
P(B/F) = P(B) P(F/A) / P(F)
= 0.4*0.01 / 0.085= 0.0471
P(C/F) = P(C) P(F/C) / P(F)
= 0.1*0.2 / 0.085= 0.2353